Triangle cross sections

If $a=b$ or $a=c$ or $b=c$ the triangle is isosceles, if $a=b=c$ the triangle is equilateral. In fact, the triangle is isosceles only when (at least) two of the lengths $a$, $b$, and $c$ are the same and it is equilateral only when three of the lengths are the same. This can be seen with a symmetry argument or by computing the side lengths of the triangle $ABC$.

By applying Pythagora's theorem to the right triangle $AVB$, the distance from $A$ to $B$ is $$\overline{AB}=\sqrt{a^2+b^2}$$ We may compute similarly the other sides: $$\overline{AC}=\sqrt{a^2+c^2} \qquad \text{and} \qquad \overline{BC}=\sqrt{b^2+c^2}$$

The law of cosines (trigonometry) allows us to compute the angles of a triangle of which the side lenghts are known. We only need to apply the formula of the law of cosines (pay attention that $a$, $b$, $c$ are here not the triangle side lengths). Alternatively, the scalar product between vectors can be used. Take $V$ as the origin and place the coordinate axes at the three cube edges from $V$, so that the points have coordinates $A=(a,0,0)$ and $B=(0,b,0)$ and $C=(0,0,c)$. The scalar product between the vectors $\overrightarrow{AB}=(-a,b,0)$ and $\overrightarrow{AC}=(-a,0,c)$ is $a^2$ hence the angle of $ABC$ at $A$ is ($\arccos$ is here the arccosinus function): $$\hat{A}=\arccos\Big(\frac{a^2}{\sqrt{a^2+b^2}\sqrt{a^2+c^2}}\Big)$$ Similarly, we have $$\hat{B}=\arccos\Big(\frac{b^2}{\sqrt{a^2+b^2}\sqrt{b^2+c^2}}\Big)\qquad \text{and} \qquad \hat{C}=\arccos\Big(\frac{c^2}{\sqrt{a^2+c^2}\sqrt{b^2+c^2}}\Big)$$

Trigonometry ensures that the area is half the product of the side lengths $\overline{AB}$ and $\overline{AC}$ times the sine of the angle at $A$. Since the sine of a triangle angle is strictly positive, we have $$\sin(\hat{A})=\sqrt{1-\cos^2(\hat{A})}$$ By substituting the value for $\cos(\hat{A})$ we find $$\sin(\hat{A})=\sqrt{1-\frac{a^4}{(a^2+b^2)(a^2+c^2)}}$$ and hence the area is $$\frac{1}{2}\overline{AB}\,\overline{AC}\sin(\hat{A})=\frac{1}{2}\sqrt{a^2b^2+a^2c^2+b^2c^2}$$ For the cube with side length $\ell$, the numbers $a$,$b$, and $c$ are most $\ell$, so the largest area is obtained when $a=b=c=\ell$ (which means that $A$, $B$, and $C$ are cube vertices). The largest possible area for the triangle $ABC$ is then $$\frac{\sqrt{3}}{2}\ell^2$$ and, for the unit cube, it is $\frac{\sqrt{3}}{2}$. Remark: The area of $ABC$ can be as small as possible, namely if $A$, $B$, and $C$ are close to $O$ the area is small and close to zero.

The triangle $ABC$ is acute. Indeed, the cosine of each internal angle of the triangle is strictly positive hence each angle is strictly less that $90^\circ$.
We show that we can obtain all acute triangles up to similarity. Without loss of generality suppose that $\overline{BC}=\sqrt{b^2+c^2}$ is the largest side, and rescale the triangle so that $\overline{BC}^2=b^2+c^2=1$. We write $$x:=\overline{AB}^2=a^2+b^2 \qquad \text{and}\qquad y:=\overline{AC}^2=a^2+(1-b^2)$$ The numbers $x$ and $y$ are strictly positive and do not exceed $1$ and their sum is greater than $1$. By varying $a$ and $b$ we can obtain all numbers $x$ and $y$ with these properties. Indeed, we can set $$a^2=(x+y-1)/2 \qquad \text{and} \qquad b^2=(1+x-y)/2$$ This shows that we can obtain all acute triangles up to similarity because the above properties for $x$ and $y$ are necessary. Firstly, it's clear that $x$ and $y$ must be strictly positive. Secondly, $x$ and $y$ cannot exceed $1$ because $\overline{BC}=1$ is the largest side length. Thirdly, the fact that $x+y$ must be greater than $1$ is related to Pythagora's theorem (and it follows from the law of cosines): in any acute triangle the sum of the squares of two sides is strictly larger than the square of the remaining side.

Up to a rescaling, we may suppose without loss of generality that we have the unit cube. With the coordinates chosen above we have the points $V=(0,0,0)$ and $A=(a,0,0)$ and $B=(0,b,0)$ and $C=(0,0,c)$. So the cube consists of the points whose coordinates are in the range from $0$ to $1$. The cutting plane consists of the points that can be obtained from $A$ by adding a multiple of the vector $\overrightarrow{AB}$ and a multiple of the vector $\overrightarrow{AC}$. These are the points $$(a,0,0)+t(-a,b,0)+s(-a,0,c)=((1-(t+s))a,tb,sc)$$ To have points of the cube, we only need to make sure that all coordinates are between $0$ and $1$. By looking at the last two coordinates, the parameters $t$ and $s$ must be non-negative and by looking at the first coordinate their sum cannot exceed $1$. Moreover, we can rewrite the above expression as $$(1-(t+s))A+t B+s C$$ The coefficients are three non-negative real numbers whose sum is precisely one. By considering the barycentrical coordinates in the triangle $ABC$ we obtain precisely all points of the triangle $ABC$. So we have proven that the intersection of the cube and the slicing plane is contained in the triangle $ABC$. To show that the triangle $ABC$ belongs to the cross section of the cube we can check that the coordinates of the above points (which are in the slicing plane) are between $0$ and $1$. Indeed, this is the case because the real numbers $a$, $b$, $c$, $(1-(t+s))$, $t$, and $s$ are between $0$ and $1$. Alternatively, we could reason by convexity because both the slicing plane and the cube are convex and contain $A$, $B$, and $C$, so they must contain the convex hull of the three points, which is the triangle $ABC$.